GATE CSE 2015 Set 3 | Question: 43

Question

The total number of prime implicants of the function $f(w, x, y, z) = \sum (0, 2, 4, 5, 6, 10)$ is __________

Comments

Make use of Quine-McCluskey Method :

 

 

 

 

 

 

 

Prime Implicants are the crossed one $(\times)$ : $m_{2}+m_{10}|m_{4}+m_{5}|m_{0}+m_{2}+m_{4}+m_{6}$

Total of 3 Prime implicants.

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answer is 3

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Quine-McCluskey Method :

not required for GATE

Answer 1

As you can see that there is  one $4$ -set and two $2$ -set that are covering the star marked $1's$ (i.e. the ones that are not covered by any other combinations).

So, the answer is $3$.

Comments

this is wrong u r saying about essential prime implecants and also EPI is 3 because where quad is formed that group must be countede as 1 so EPI is also 3

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This is Essential Prime Implicants not Prime Implicants.

Prime Implicants is all possible combinations right? I'm getting 6

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No, $PI = 3$ correct.

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doesn't prime implicants mean the number of product terms that come in the minimized sum of products....

correct me if i am wrong

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No, Not necessary.

Prime Implicant : the biggest subcube that should not be completely covered by any other subcube (some may be covered) 

Have a look my this answer. ( this can be of some help to you)

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then what is the difference between PI and EPI?

Here is what wiki says :

As per this definition PI may have all it's outputs covered, but EPI shouldn't

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@Shalini EPI are marked with * in the given answer. And question asks for no. of PI and I guess you are counting no. of distinct terms which can come in any PI.

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Yes Sir. But I'm failing to understand the definition of PI. If a ques asks for num of PIs, then we should consider minimal covering of the map?(i.e. 1 subcube should not be completely covered by another subcubes, for all subcubes)

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Yes, you get one minimal cover. Now, try to change this- how many changes you can do is the answer.

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PI is three and Is number of essential prime implicant 2?

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@shivangi EPI are term which are not covered and they are 4 marked by *

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EPI should be 3 not 4.

As we are able to cover all 1's in 3 PI.

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I agree that no of EPI = 3

Answer 2

(Can be solved using K-Map also. )

Place all minterms that evaluate to one in a minterm table.

Input (first column for no. of 1's)

0	m0	0000
1	m2 m4	0010 0100
2	m5 m6 m10	0101 0110 1010

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Combine minterms with other minterms. If two terms vary by only a single digit changing, that digit can be replaced with a dash indicating that the digit doesn't matter. Terms that can't be combined any more are marked with a "*". When going from Size 2 to Size 4, treat '-' as a third bit value. For instance, -110 and -100 or -11- can be combined, but -110 and 011- cannot. (Trick: Match up the '-' first.)

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First Comparison

0	(2, 0) (4, 0)	00-0 0-00
1	(6, 2) (10, 2) (5, 4) (6, 4)	0-10 -010 010- 01-0

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Second Comparison

0

(6, 4, 2, 0)

0--0

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Prime Implicants

(6, 4, 2, 0)	0--0
(10, 2) (5, 4)	-010 010-

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Answer: Total number of prime implicants 

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Source: Finding prime implicants - Quine-McCluskey algorithm - Wikipedia

Comments

is Quine McCluckey algo in gate syllabus

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na ,not in syllabus

Answer 3

all the min terms can only be covered by three diffrernt color so there is only three prime imlecants

Answer 4

I think this is bit easier to understand.
The hint here is that we try to find the minimum number of groups(octa(eight 1's) ,quad (four 1's),dual(two 1's) that can be formed and that represents the minimum number of implicants that covers F which is Prime Implicant.

Here, we can form 1 quad and 2 duets.3 prime implicants.

 

Comments

The blue combination is wrong. There is two bits of difference between the blue combination.

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@Harsh Kumar-Yes, this is wrong. Thanks :)

Corrected Now. Please verify.